Question

A particle moves along $x-$axis in positive direction. Its acceleration a is given as $a=cx+d$, where $x$ denotes the $x-$coordinate of particle, $c$ and $d$ are positive constants. For velocity-position graph of particle to be of type as shown in figure, the value of speed of particle at $x=0$ should be:

• A. $\sqrt{\frac{4d^2}{c}}$
• B. $\sqrt{\frac{d^2}{c}}$
• C. $\sqrt{\frac{2d^2}{c}}$
• D. $\sqrt{\frac{8d^2}{c}}$

Answer 1

The correct option is B $\sqrt{\frac{d^2}{c}}$

$a=cx+d$

$u=kx+u$..(1)

$u.\frac{dv}{dx}=cx+d$

$vk=cx+d$

$v=\frac{kcx}{k^2}+\frac{d}{k^2}$...(2)

Compare equation (1) and (2)

$k=\sqrt{c}$

$u=\frac{d}{k}=\frac{d}{\sqrt{c}}$